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anonymous

  • 5 years ago

dos the system have on solution, no solution, or an infinite number of solutions? x+y=6 x+y=4 a. one. b. none. c. infinite number

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  1. amistre64
    • 5 years ago
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    none, x+y can only have one solution..otherwise: 4+5 = 9 4+5 = 3 just nonsense i tell ya....nonsense!! :)

  2. anonymous
    • 5 years ago
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    lol thank yiu

  3. amistre64
    • 5 years ago
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    graphically, and if it aint a word it should be, these are parallel lines that never intersect

  4. radar
    • 5 years ago
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    How about this one? x + y =4 x + y =0 Would x=2\[y=\sqrt{4}\] be an answer????

  5. amistre64
    • 5 years ago
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    same slopes, different "y intercepts"....nope

  6. amistre64
    • 5 years ago
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    y=-x y = -x+4 -x = -x+4 -x+x = 4 0 = 4 nonsense :)

  7. radar
    • 5 years ago
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    Just wondering, as the square root of 4 is +2 or -2 and they when substituted for y and if x=2 would produce those two answers. Kind caught me off guard.

  8. amistre64
    • 5 years ago
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    2+2 = 4 2+2 = 0 still not a good solution :)

  9. anonymous
    • 5 years ago
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    Square root of 4 is +2 (by definition), kids [NOT -2]

  10. amistre64
    • 5 years ago
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    even if -2 was available to us it doesnt change the results of 0=4; which is absurd....

  11. anonymous
    • 5 years ago
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    He was implying it could equal 2 or -2 (one in each) - yes, of course that is absurd, but I thought I would tell him the flaw in his thinking.

  12. amistre64
    • 5 years ago
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    yeah, 2 = -2 is equally nonsensical :) If we cant pin down x at a single value, then its futile.

  13. radar
    • 5 years ago
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    Thanks, you can see where the flaw would 2 +(+2)=4 and 2 + (-2) = 0 However, it has been revealed that the negative root of a variable is not usable or shouldn't be used in this case. Now please note. There was no way to solve the two equations to come up with that ridiculous answer!

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